Cauchy criterion for convergence

If a sequence converges, it is the same as saying it matches the Cauchy criterion for convergence.

Cauchy Sequence

A sequence

$$(a_n)^\infty_{n=1}$$ is Cauchy if:

$$\forall\epsilon>0\exists N\in\mathbb{N}:n> m> N\implies d(a_m,a_n)<\epsilon$$

Theorem

A sequence converges if and only if it is Cauchy

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TODO: proof, easy stuff

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Interesting examples

$$f_n(t)=t^n\rightarrow 0$$ in $$\|\cdot\|_{L^1}$$

Using the

$$\|\cdot\|_{L^1}$$ norm stated here for convenience: $$\|f\|_{L^p}=\left(\int^1_0|f(x)|^pdx\right)^\frac{1}{p}$$ so $$\|f\|_{L^1}=\int^1_0|f(x)|dx$$

We see that

$$\|f_n\|_{L^1}=\int^1_0x^ndx=\left[\frac{1}{n+1}x^{n+1}\right]^1_0=\frac{1}{n+1}$$

This clearly

$$\rightarrow 0$$ - this is $$0:[0,1]\rightarrow\mathbb{R}$$ which of course has norm $0$, we think of this from the sequence $$(\|f_n-0\|_{L^1})^\infty_{n=1}\rightarrow 0\iff f_n\rightarrow 0$$

[Expand]Real Analysis

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Functional Analysis

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